ECE302 Spring 2015 HW5 Solutions March 27, 2015 1

Solutions to Homework Assignment 5

Problem 3.6.3 and Solution

Given X has CDF

FX(x) =

0 x < −1

0.2 x ∈ [−1, 0)0.7 x ∈ [0, 1)

1 x ≥ 1

(1)

(a) Find PV (v) where V = g(X) = |X|.

First, we find PX(x) to be

PX(x) =

0.2 x = −1 (0.2− 0)0.5 x = 0 (0.7− 0.2)0.3 x = 1 (1− 0.7)

0 otherwise

(2)

by subtracting the value of FX(x) at the lower edge of a discontinuity fromFX(x) at the upper edge of the discontinuity. Then from its definition,

SV = {0, 1}

and

PV (v) =

0.5 v = 1 (0.2 + 0.3)0.5 x = 0

0 otherwise.(3)

(b) Find FV (v).

FV (v) =

0 v < 0

0.5 0 ≤ v < 11 v ≥ 1

(4)

(c) Find E[V ].

E[V ] =∑SV

vPV (v) = 1(1/2) + 0(1/2) = 1/2. (5)

ECE302 Spring 2015 HW5 Solutions March 27, 2015 2

Problem 3.6.4

Given: The transaction fee f for a stock purchase d is

f(d) =

{d0.01de d ≤ 10, 000

d100 + 0.005(d− 10, 000)e d > 10, 000(6)

where we are redefining the ceiling function to indicate rounding up to thenearest cent.

Consider stock with price per share s ∈ {99.75, 100, 100.25} having PMF

PS(s) =

{1/3 p ∈ {99.75, 100, 100.25}

0 otherwise.(7)

Find the PMF of the cost C of 100 shares.

Problem 3.6.4 Solution

The cost isC = d + f(d) where d = 100s.

We then calculate the fees for each possible cost of the block of 100 shares.

d = 100(99.75) = 9975 f(d) = .01(100)99.75 = 99.75

d = 100(100) = 10, 000 f(d) = .01(100)100 = 100

d = 100(100.25) = 10, 025 f(d) = d100 + 0.005(10025− 10000)e= d100.125e = 100.13

(8)

Thus

PC(c)

{1/3 c ∈ {10074.75, 10100, 10125.13}

0 otherwise.(9)

skoskie

Rectangle